Question

Identify the correct statement(s):

• A. The oxidation number of Cr in CrO₅ is +6.
• B. ΔH>ΔU for the reaction N₂O₄(g) → 2 NO₂(g), provided both gases behave ideally.
• C. pH of 0.1 N H₂SO₄ is less than that of 0.1 N HCl at 25°C.
• D. RT/F = 0.0591 V at 25°C.

Hint:

Use \(\Delta H = \Delta U + \Delta nRT \) to compare enthalpy and internal energy. For pH calculations, consider the dissociation capability of acids.

Answer 1

The Correct Option is A, B

Analysis of Each Statement:

• (A) The oxidation number of Cr in CrO₅ is +6.
• CrO₅ (chromium pentoxide) is a peroxide compound with one Cr atom and five O atoms. To find Cr's oxidation number:

\( \text{Oxidation number of Cr} + 5(-2) = 0 \).

• Since 4 oxygen atoms form peroxide bonds (each O is -1), the calculation is:

\( \text{Oxidation number of Cr} + (-2 \times 4) + (-2) = 0 \)

• Solving yields:

\(\text{Oxidation number of Cr} = +6\).

• This statement is correct.
• (B) ΔH > ΔU for the reaction N₂O₄(g) → 2 NO₂(g), provided both gases behave ideally.

For an ideal gas:

\( \Delta H = \Delta U + \Delta n_gRT \),

• where \(\Delta n_g\) is the change in moles of gas. For the reaction:

\(\text{N}_2\text{O}_4(g) \rightarrow 2\text{NO}_2(g)\),

\( \Delta n_g = 2 - 1 = 1\). Therefore, \(\Delta H > \Delta U\). This statement is correct.

• (C) pH of 0.1 N H₂SO₄ is less than that of 0.1 N HCl at 25°C. H₂SO₄ fully dissociates into 2 H⁺ and 1 SO²⁻₄ per molecule. However, in 0.1 N, the second ionization of H₂SO₄ is suppressed, leading to slightly more than 0.1 M H⁺. 0.1 N HCl gives exactly 0.1 M H⁺. Thus, the pH of 0.1 N H₂SO₄ is slightly higher (less acidic) than 0.1 N HCl. This statement is incorrect.
• (D) RT/F = 0.0591 V at 25°C. The RT/F relation at 25°C is:

\( \frac{RT}{F} = \frac{(8.314)(298)}{96485} \approx 0.0257 \text{ V}. \)

• The value 0.0591 V corresponds to \(2.303 \times \frac{RT}{F}\), used in the Nernst equation for logarithmic expressions. This statement is incorrect.

Conclusion: The correct statements are: (A), (B).